Download Solutions

MTHE/STAT 351 — Fall 2016:
Solutions to Homework Assignment 2
1. [A real number x is selected. . . ]
(a) The sample space S is the interval [−2, 3]. Since x is always in this interval we can express the
events A, B, and C as subsets of [−2, 3]:
A = {x ∈ [−2, 3] : x < − 12 } = [−2, − 12 )
B = {x ∈ [−2, 3] : |x| < 1} = [−2, 3] ∩ (−1, 1) = (−1, 1)
C = {x ∈ [−2, 3] : x ≥ 21 } = [ 21 , 3]
Now recall that is a point is chosen randomly from [−2, 3], then for any −2 ≤ α ≤ β ≤ 3 we have
P ([α, β]) =
β−α
β−α
=
3 − (−2)
5
Also recall that since P ({x}) = 0 for any single point x,
P ([α, β]) = P ((α, β]) = P ([α, β)) = P ((α, β))
Thus we have
P (A) =
P ([−2, − 12 ))
− 12 − (−2)
3
=
=
5
10
and similarly
2
P (B) = P (−1, 1) =
5
1
P (C) = P ([ 12 , 3]) =
2
Moreover
1
P (AB) = P [−2, − 12 ) ∩ (−1, 1) = P (−1, − 12 ] =
10
1
1
P (AC) = P [−2, − 2 ) ∩ [ 2 , 3] = P (∅) = 0
and since
A − B = [−2, − 12 ) − (−1, 1) = [−2, −1]
and
(A − B) − C = [−2, −1] − [ 12 , 3] = [−2, −1]
we have
1
P (A − B) − C = P ([−2, −1]) = .
5
1
(b) The solution of x2 − x = 2 are x1 = 2 and x2 = −1. Thus we want to find the probability of
the set {−1, 2}. Using the additivity of the probability
P ({−1, 2}) = P ({−1} ∪ {2}) = P ({−1}) + P ({2}) = 0
since P ({x}) = 0 for any x ∈ [−2, 3]
2. [Ghahramani, 2.2, # 18]
To count the number of 4-digit extensions with no repeated digits, we break the counting task into
4 subtasks, where subtask j is to count the number of ways to choose the jth digit in the sequence.
For the first digit we may choose one of 10 values. Once the first digit is chosen we may choose the
second digit from one of the remaining 9 values. Similarly, the third digit can be one of 8 values
and the fourth digit can be one of 7 values. So the number of ways to choose a 4-digit number with
no digits repeating is 10 × 9 × 8 × 7 = 5040.
(a) To count the number of 4-digit numbers with distinct digits that do not start with a 0, it is
simpler to count the number that do start with a 0. The second digit can be one of 9 values,
then the third digit can be one of 8 values, then the fourth digit can be one of 7 values. So
there are 9 × 8 × 7 = 504 4-digit numbers with distinct digits that do start with a 0 and
5040 − 504 = 4536 4-digit numbers with distinct digits that do not start with a 0.
(b) Proceeding as above, the number of 4-digit numbers with distinct values that do start with 01
is 8 × 7 = 56, so the number that do not start with 01 is 5040 − 56 = 4984.
3. [Ghahramani, 2.2, # 23]
Let S denote the set of all integers between 1 and 1,000,000. Since all elements of S are equally
likely to be selected, we have to count how many of them contain the digit 5 in their usual decimal
representation. Furthermore, it is easier to count how many of them do not contain the digit 5. It
will be useful to represent all elements of S except 1,000,000 by the following set of 6-digit decimals
B = {d6 d5 d4 d3 d2 d1 : 0 ≤ di ≤ 9, i = 1, . . . , 6} \ {000000}
(thus, for example, 123 is represented as 000123). Then
S = B ∪ {1000000}.
If A is the subset of decimals in S that contain the digit 5, then
Ac = {1000000} ∪ {d6 d5 d4 d3 d2 d1 : 0 ≤ di ≤ 9, di 6= 5, i = 1, . . . , 6} \ {000000},
Clearly, the set in the middle has 96 elements. We obtain |Ac | = 1 + 96 − 1 = 96 . Therefore
P (A) = 1 − P (Ac ) = 1 −
2
|Ac |
96
= 1 − 6 = 0.469.
|S|
10
4. [A child has 12 blocks . . . ]
(a) The number of ways the blocks can be arranged in a line is the number of distinguishable
permutations of 12 objects of 4 different types such that 6 are type 1 (black), 4 are type 2
(red), 1 is type 3 (white), and 1 is type 4 (yellow). The number of all such distinguishable
permutations is
12!
= 27720.
6!4!1!1!
(b) Since there are 12 blocks, six of which are black, the condition that no two black blocks are
neighbors means that between every two consecutive black blocks there is at least one block of
a different color. The line either starts with a black block or with a block of different color. If
the line starts with a different color, then between two black blocks there must be exactly one
block of different color. If the line starts with a black block, then either black and different
colors alternate (and the line ends with a different color), or there exist exactly two black
blocks between which there are exactly two blocks of different color, and in the rest of the
line the black and differently colored blocks alternate (the line ends with a black block). In
the latter case, there are 5 ways of choosing where to insert the two blocks of different color
between the two black blocks. Therefore, if we consider the non-black blocks indistinguishable,
there are 7 different arrangements such that no two black blocks are next to each other. In all
7 cases, the 6 non-black blocks (now viewed as 4 red, 1 white, and 1 yellow) can be arranged
in
6!
= 30
4!1!1!
different ways. Thus the number of arrangements in question is 7 · 30 = 210 and the desired
probability is
210
≈ 0.00757.
27720
5. [Review problem 16, p. 73]
The sample space S is the set of all possible arrangements of four women and two men at a table
with four seats on one side and four on the opposite side. Since all seating arrangements are equally
likely, we will have to count the number of arrangements such that not all men are sitting on one
side. We will count the number of elements in the complementary event (all men are sitting on one
side), not because it is simpler to do that in this particular case, but because this may lead to a
simpler solution in more general cases (e.g., if there are 4 men, 4 women, and 5 seats on both sides).
In counting the seating arrangements, we will consider all women indistinguishable and all men
indistinguishable. Suppose the seats are numbered from 1 to 8, such that seats 1,2,3, and 4 are
on one side, the rest on the other side. A given seating arrangement is uniquely represented by
the (unordered set of) two seat numbers where the men sit, and the (unordered set of) four seat
3
numbers where the women sit (for example ({1, 4}, {2, 5, 6, 8}) is the seating arrangement where
the two men sit on seats 1 and 4 and the women sit on seats 2, 5, 6, and 8). The number of all
possible arrangements is
8 6
|S| =
2 4
8
since there are 2 ways for the two (indistinguishable) men to choose two of the possible eight
seats, and 64 ways for the four (indistinguishable) women to choose four of the remaining six seats.
Similarly, letting A denote the arrangements where the two men sit on different sides, we have
4 6
c
|A | = 2
2 4
since there are 2 42 ways for the two men to choose two seats on the same side of the table ( 42
ways on each side), and 64 ways for the four women to choose four of the remaining six seats. Thus
we get
4 6
2
c
4
|A |
2 4
= = = 0.571.
P (A) = 1 − P (Ac ) = 1 −
8 6
|S|
7
2 4
6. [Consider a 4 × 6 grid . . . ]
(a) Note that each allowed path must consist of 10 steps, 4 of which must be “ups”, and the remaining 6 must be “rights”. Thus, paths of the specified kind correspond precisely to the distinct
permutations of 4 U’s (“ups”) and 6 R’s (“rights”). The total number of such permutations,
10!
and hence the total number of up/right paths from (0,0) to (6,4), is 4!6!
= 10
= 210.
4
(b) An up/right path from (0,0) to (6,4) passes through the point (3,2) if and only if the first 5
steps of the path consist of 2 U’s and 3 R’s, and the last 5 steps also consist of 2 U’s and 3
5!
5!
R’s. Hence, the number of paths from (0,0) to (6,4) that pass through (3,2) is 2!3!
× 2!3!
= 100.
So, the required probability is
100
210
=
10
21
≈ 0.476.
Bonus question [We are given n points. . . ]
Let P denote the set of n points. Since no four points are co-planar, any three-element subset of
P uniquely determines a plane in Ω (in fact, Ω can be identified with the collection of three-element
subsets of P ). A straight line that is the intersection of two planes in Ω passes through a point
in P if and only if that point belongs to both planes, i.e., the point belongs to both three-element
subsets determining the two planes. Thus we have to count the number N0 of (unordered) pairs of
planes in Ω with no points in P in common, as well as the number N of all (unordered) pairs of
planes in Ω.
4
Clearly,
1 n n−3
N0 =
2 3
3
and
n
3
N=
2
.
Thus the desired probability is
1 n n−3
n−3
N0
2 3
3
3
= = .
n
1 n
n
N
−1
−1
3
2 3
3
Note that we can further simplify this result as follows:
n−3
(n − 4)(4 − 5)
3
=
.
n
n2 + 2
−1
3
5